Brain teasers and challenges.

[Ohm]

Firstly, I will prove that 1 = 2 and then prove that four added to six will make eleven... [a2 and b2 means a square and b square respectively]

1. Let a = b
2. a2 = ab (Multiplying both sides by a)
3. a2 - b2 = ab - b2 (Subtract b2 from both sides to get)
4. So, this implies that (a + b)(a - b) = b(a - b)
5. a + b = b (Since (a - b) appears on both sides, we can cancel)
6. b + b = b (Since a = b)
7. 2b = b
8. So we get that, 2 = 1 or 1 = 2 (cancelling b on both sides)

Now, add '9' on both sides
This gives us, 10 = 11
6 + 4 = 11 (10 can be broken into 6 + 4)
Hence Proved....

In third step, it is given that a^2-b^2 = ab-b^2. It also can be written as a^2-a^2 = a*a-a^2 (since a=b), so the simplification ends as 0 = 0. You mathematically proved 1=2, but it failed even mathematically. Very good attempt anyways.

It says 'show' four added to six will make eleven. There's a small hint. You don't always need to be algebraic for that.

 

[Abhas]

I was thinking Roman Numerals, but doesn't make much sense to me..
IV + VI = 11 / ELEVEN / XI

 

[Ohm]

I was thinking Roman Numerals, but doesn't make much sense to me..
IV + VI = 11 / ELEVEN / XI

That's correct thinking. The answer can be generated graphically using IV and VI.



By the way puzzles are invented for reasons, otherwise we always have math to solve or simplify. That puzzle was based on numerical and some graphics so the logic makes sense imo.

 

[Pinch hitter]

Diff: Medium
Can you show that four added to six will make eleven?

i think answer is 4+6=11

 

[Ohm]

i think answer is 4+6=11

That's high order intelligence, and I've nothing much to add about your headpiece.

 

[Pinch hitter]

Diff: Easy
How John could find the honest and liar?

I think John find that both are honest... When John asked 1st person, he said Left and and 2nd person said right, but he find out that 2nd person not telling right direction... He is telling that, the 1st person told direction is right

 

[Ohm]

I think John find that both are honest... When John asked 1st person, he said Left and and 2nd person said right, but he find out that 2nd person not telling right direction... He is telling that, the 1st person told direction is right

Humbly I think you didn't get this rightly(correctly): "When John asked them on driving which turn D comes, one of them answered left and other one answered right." They both answered to John's question but the different words.

They just said 'left' or 'right'. Just like,

John: How shall I reach to D?
1st person: Take right turn.
2nd person: Take left turn.

But not both are honest or both liar either. One of them is honest and other one is liar as described.

 

[lion100lion]

In third step, it is given that a^2-b^2 = ab-b^2. It also can be written as a^2-a^2 = a*a-a^2 (since a=b), so the simplification ends as 0 = 0. You mathematically proved 1=2, but it failed even mathematically. Very good attempt anyways.

I have changed the proof of 1=2 :-

1 square = 1
2 square =4=2+2
3 square=9=3+3+3
4 square=16=4+4+4+4
5 square=25=5+5+5+5+5
.
.
.
.
.
.
.
.
.
Similarly 'X' square = X + X + X +............ (X times)
Now, when you differentiate both sides w.r.t X, you get
2X = 1 + 1 + 1 +1 +......... (X times)
which implies that 2X = X
So, 2=1 or 1=2 (X cancelled on both sides)



The rest of procedure is same as in my above post :-

Now, add '9' on both sides
This gives us, 10 = 11
6 + 4 = 11 (10 can be broken into 6 + 4)
Hence Proved....

 

[ethybubs]

What if he asks that question to the liar, he/she would suggest exactly the correct path in reply

No they wouldn't, they would say the incorrect path because the honest person would point to the correct path, but because the liar always lies, he would say the wrong path and therefore I can ask the question to either one and know the right path.

If you are going to ask a riddle, at the very least know the answer.

 

[Ohm]

No they wouldn't, they would say the incorrect path because the honest person would point to the correct path, but because the liar always lies, he would say the wrong path and therefore I can ask the question to either one and know the right path.

If you are going to ask a riddle, at the very least know the answer.

That wasn't the answer I expected or I knew that should be the reason I didn't get it on reading earlier. Your answer was correct anyways for some possibilities, means only if both person know about each other's character and in some case they will know during the conversation.
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My actual answer - Ask both a common question "How shall I reach to A?" One would point out the correct path(honest) but other would tell the path that take him either to D or B. Since John already knew the correct answer, now the one speaking truth and one who lie have been found. Follow honest's advice.

 

[Ohm]

Math things:

A man left $100.00 to be divided between his two sons Alfred and Bruce. If one third of Alfred's legacy be taken from one-fourth of Bruce's legacy, the remainder would be $11.00. What was the amount of each legacy?

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Can you find two numbers composed of ones which give the same result by addition and multiplication?

 

[Abhas]

Can you find two numbers composed of ones which give the same result by addition and multiplication?

II+II = IV
II x II = IV

 

[Ohm]

^Hehehe I knew somebody would say that But can you do that using '1' or/and '1s'?

 

[Abhas]

A man left $100.00 to be divided between his two sons Alfred and Bruce. If one third of Alfred's legacy be taken from one-fourth of Bruce's legacy, the remainder would be $11.00. What was the amount of each legacy?

Can't explain it any better than this..

x+y=100

x/3 + 11 = y/4

y = 4x/3 +44

x + 4x/3 + 44 = 100

(3x+4x)/3=56
7x=168
x = 24

Alfred has 24, Bruce has 76

 

[ethybubs]

Can you find two numbers composed of ones which give the same result by addition and multiplication?

11*1.1 = 12.1
11+1.1 = 12.1